While finding out the most probable location from wavefunction, why don't we set the derivative of probability to zero?

The probability density function is what you need to differentiate, not the integral. $$P = int_D |psi|^2 dx = 1$$ for any normalized state for $D$ the domain of the problem. So you have to look at the probability density, $f(x) = |psi(x)|^2$, and find its maximum. More formally, the probability of finding the particle described by $psi(x)$ within the infinitesimal interval $(x, x+dx)$ is given by $$int_x^{x+dx} f(x)dx = |psi(x)|^2 dx $$

Finding the most probable location, amounts to finding the maximum of $f(x)$ for which you can take derivative and equate it to zero, then solve for $x$. Notice that you technically require a non-zero volume (could be small, but not zero) to actually obtain a non-zero probability. Units make sense therefore, in one dimension $$text{probability} = frac{text{probability}}{text{lenth}}cdot text{length} = |psi(x)|^2 Delta x$$

P.D. Don't confuse with the expected value of the position or, the average of the position. $$langle psi | hat{x} | psi rangle = int x|psi(x)|^2dx$$.